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I am trying to derive 4-velocity, but i get stucked when i try to derive equation for proper time. I end up confused as i get $\scriptsize\textrm{d}\tau = \frac{\textrm{d}t}{\gamma(v)}$ instead of $\scriptsize \textrm{d}\tau = \frac{\textrm{d}t}{\gamma}$. Is this even correct? I am doing this for a proper coordinate system where $\scriptsize \Delta x=\Delta y=\Delta z = 0$.

\begin{equation} \scriptsize \begin{split} \Delta s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2 - (c \Delta t)^2 &= 0 + 0 + 0 - (c \Delta \tau)^2\\ \textrm{d} x^2 + \textrm{d} y^2 + \textrm{d} z^2 - (c \textrm{d} t)^2 &= 0 + 0 + 0 - (c \textrm{d} \tau)^2\\ \frac{\textrm{d} x^2 + \textrm{d} y^2 + \textrm{d} z^2}{c^2} - \textrm{d} t^2 &= -\textrm{d} \tau^2\\ \frac{(v_x \textrm{d} t)^2 + (v_y \textrm{d} t)^2 + (v_z \textrm{d} t)^2}{c^2} - \textrm{d} t^2 &= -\textrm{d} \tau^2\\ \frac{v^2 \textrm{d} t^2}{c^2} - \textrm{d} t^2 &= -\textrm{d} \tau^2\\ \textrm{d}\tau^2 &= \textrm{d} t^2 \frac{v^2 \textrm{d}t^2}{c^2}\\ \textrm{d}\tau^2 &= \textrm{d} t^2 \left(1 - \frac{v^2}{c^2}\right)\\ \textrm{d}\tau &= \frac{\textrm{d} t}{\gamma(v)} \end{split} \end{equation}

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You get A instead of B, we hear. The only difference between A and B is that in A, the dependence of $\gamma$ on $v$ is explicitly written down. What's the difference? The derivation is right although a bit long. –  LuboŇ° Motl Dec 10 '12 at 12:11
So this is all ok? –  71GA Dec 10 '12 at 12:29
It is OK except that the 4-velocity, a 4-vector, from the title, never appears in the derivation; and except that your question about the disagreement that only boils down to the missing $(v)$ indicates that the derivation may not be yours because otherwise you would know that it's up to you whether you write $(v)$ or not... You know, if it's a homework and you copied the derivation from somewhere, the derivation may be a good answer but it may be the answer to a different question. ;-) –  LuboŇ° Motl Dec 10 '12 at 13:32
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